Question

13. [-/5 Points] DETAILS MY NOTES Using the Integral Test on the series $\sum_{n=0}^{\infty} \frac{n}{(n^2 + 92)^2}$ we see that the integral $\int_{0}^{\infty} \frac{x}{(x^2 + 92)^2} dx$ $\circ$ converges to $\frac{1}{46}$ $\circ$ diverges $\circ$ converges to $\frac{1}{184}$ $\circ$ converges to 0 $\circ$ converges to $\frac{1}{92}$ and hence the series $\sum_{n=0}^{\infty} \frac{n}{(n^2 + 92)^2}$ is $\circ$ convergent $\circ$ divergent Submit Answer

          13. [-/5 Points]
DETAILS
MY NOTES
Using the Integral Test on the series $\sum_{n=0}^{\infty} \frac{n}{(n^2 + 92)^2}$ we see that the integral $\int_{0}^{\infty} \frac{x}{(x^2 + 92)^2} dx$
$\circ$ converges to $\frac{1}{46}$
$\circ$ diverges
$\circ$ converges to $\frac{1}{184}$
$\circ$ converges to 0
$\circ$ converges to $\frac{1}{92}$
and hence the series $\sum_{n=0}^{\infty} \frac{n}{(n^2 + 92)^2}$ is
$\circ$ convergent
$\circ$ divergent
Submit Answer

13. [-/5 Points]
DETAILS
MY NOTES
Using the Integral Test on the series ∑n=0^∞(n)/((n^2 + 92)^2) we see that the integral ∫0^∞(x)/((x^2 + 92)^2) dx
∘ converges to (1)/(46)
∘ diverges
∘ converges to (1)/(184)
∘ converges to 0
∘ converges to (1)/(92)
and hence the series ∑n=0^∞(n)/((n^2 + 92)^2) is
∘ convergent
∘ divergent
Submit Answer

Added by Megan A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Vincenzo Zaccaro

05/13/2023

Step 1

Step 1: The Integral Test states that if f(x) is a positive, continuous, and decreasing function on the interval [1, ∞) and if the series Σ a_n is given by a_n = f(n), then the series Σ a_n converges if and only if the improper integral ∫ f(x) dx from 1 to ∞ Show more…

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